Question

In the reaction
BrO₃⁻ (aq) +5Br⁻ (aq) + 6H⁺ (aq) 3Br₂(l) +3H₂O(l) - ®
The rate of appearance of bromine (Br₂) is related to rate of
disappearance of bromide ions as following:
(a) d[Br₂ ]/dt = – 5/3 d[Br⁻]/dt
(b) d[Br₂ ]/dt = 5/3 d[Br⁻]/dt
(c) d[Br₂ ]/dt = 3/5 d[Br⁻]/dt
(d) d[Br₂ ]/dt = – 3/5 d[Br⁻]/dt

Answer 1

answer : option (d) d[Br₂ ]/dt = – 3/5 d[Br⁻]/dt

we know,

rate of appearance or disappearance = ± 1/stoichiometric coefficient × d[product or reactant]/dt

for instance, aA + bB ⇒cC + dD

-1/a d[A]/dt = -1/b d[B]/dt = +1/c d[C]/dt = +1/d d[D]/dt

now come to the question, reaction is ....

BrO₃⁻ (aq) +5Br⁻ (aq) + 6H⁺ (aq) + 3Br₂(l) +3H₂O(l)

so, -1/5 d[Br⁻]/dt = +1/3 d[Br₂]/dt

⇒d[Br₂]/dt = -3/5 d[Br⁻]/dt

hence option (d) is correct choice.

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Answer 2

Answer will be (d)

Explanation:

Here,you are asking for rate of appearance of bromine($Br_{2}$) which is produced by bromide ion($Br^{-}$) here

And rate of appearance or disappearance  of bromide

$d[Br_{2}]/dt$÷(stoichiometric coefficient)= $+d[Br^{-}]/dt$÷(stoichiometric coefficient)

(+) sign for product

(-) sign for reactant

so here ,$-(1/5)d[Br_{2}]/dt$$= +(1/3)d[Br^{-}]/dt$

$d[Br_{2}]/dt= -3/5d[Br^{-}]/dt$

option (d) is the answer